给定的质数分解遍历c ++中的所有因子而无需递归

Question

I am given the prime factorization of a number p1^x1 * p2^x2 * .... in a map. I need to iterate through all its factors, prime as well as composite. I managed to write a solution using recursion.

#include <iostream>
#include <map>
#include <cstdlib>

using namespace std;

struct PROBLEM {

    int mx = 400;
    map<int, int> mp = {{2, 2}, {3, 1},  {5, 1}, {7, 2}};
    int lastPrimeFactor = 7;
    int num = 1;

    auto solve() {
        rec(2, 0);
        return 0;
    }

    int next_prime_factor(int p) {
        return (p == 2) ? 3 : (p == 3) ? 5 : (p == 5) ? 7 : -1;
    }

    void rec(int prime, int power) {

        if (mx == 0) {
            cout << "Infinite recursion\n\n";
            exit(0);
        } else --mx;

        if (prime == lastPrimeFactor && power > mp[prime]) {
            return;
        }

        if (power < mp[prime]) {
            num *= prime;
            cout << num << endl;
            rec(prime,  power + 1);
            num /= prime;
        }

        if (prime != lastPrimeFactor) {
            rec(next_prime_factor(prime),  0);
        }

    }

};


int main() {
    PROBLEM().solve();
    return 0;
}

Questions:

1) Is there any faster way to generate these factors?

2) If possible, can I replace the recursion by a while loop?

Answer 1

1. No . Your recursive algorithm works in exactly the same time as the number of divisors. Any algorithm which works asymptotically faster cannot print all these numbers.

2. Yes . Any recursive algorithm may be rewritten in a non-recursive way using the std::stack to store local variables. But, in your case this will not probably be faster and will make the code much less readable, so such rewrite is undesirable. If necessary, I can provide you code.

Answer 2

Without recursion, it may look like:

bool increase(const std::vector<std::pair<std::size_t, std::size_t>>& v,
              std::vector<std::size_t>& it)
{
    for (std::size_t i = 0, size = it.size(); i != size; ++i) {
        const std::size_t index = size - 1 - i;
        ++it[index];
        if (it[index] > v[index].second) {
            it[index] = 0;
        } else {
            return true;
        }
    }
    return false;
}

std::size_t pow(std::size_t n, std::size_t power)
{
    std::size_t res = 1;
    for (std::size_t i = 0; i != power; ++i) {
        res *= n;
    }
    return res;
}

void do_job(const std::vector<std::pair<std::size_t, std::size_t>>& v,
            std::vector<std::size_t> it)
{
    std::size_t res = 1;
    for (std::size_t i = 0; i != v.size(); ++i) {
        res *= pow(v[i].first, it[i]);         
    }
    std::cout << res << std::endl; 
}

void iterate(const std::vector<std::pair<std::size_t, std::size_t>>& v)
{
    std::vector<std::size_t> it(v.size(), 0);

    do {
        do_job(v, it);
    } while (increase(v, it));
}

Demo

So basically, we count from {0, 0, 0, 0} to {2, 1, 1, 2} .